A satellite is orbiting at a certain height in a circular orbit. If the mass of the planet is reduced to half the initial value, then the satellite would

Options

A. fall on the planet
B. go to the orbit of smaller radius
C. go to the orbit of larger radius
D. escapes from the planet

Answer

To solve the problem of what happens to a satellite when the mass of the planet is reduced to half, we can follow these steps:

### Step 1: Understand the initial conditions
The satellite is initially orbiting a planet of mass \( M \) at a certain height. The gravitational force provides the necessary centripetal force for the satellite to maintain its circular orbit.

### Step 2: Write the formula for orbital velocity
The orbital velocity \( v_0 \) of a satellite in a circular orbit is given by the formula:
\[ v_0 = \sqrt{\frac{GM}{r}} \]
where:
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the planet,
- \( r \) is the distance from the center of the planet to the satellite.

### Step 3: Determine the new conditions after reducing the mass
If the mass of the planet is reduced to half, the new mass \( M' \) becomes:
\[ M' = \frac{M}{2} \]

### Step 4: Write the new formula for orbital velocity
The new orbital velocity \( v_0' \) with the reduced mass is:
\[ v_0' = \sqrt{\frac{G \cdot M'}{r}} = \sqrt{\frac{G \cdot \frac{M}{2}}{r}} = \sqrt{\frac{GM}{2r}} = \frac{v_0}{\sqrt{2}} \]

### Step 5: Compare the new orbital velocity with escape velocity
The escape velocity \( v_e \) from the planet is given by:
\[ v_e = \sqrt{\frac{2GM}{r}} \]
For the new mass:
\[ v_e' = \sqrt{\frac{2G \cdot M'}{r}} = \sqrt{\frac{2G \cdot \frac{M}{2}}{r}} = \sqrt{\frac{GM}{r}} = v_0 \]

### Step 6: Conclusion
Since the new escape velocity \( v_e' \) is equal to the original orbital velocity \( v_0 \), this means that the satellite's velocity is now equal to the escape velocity. Therefore, the satellite will escape the planet.

### Final Answer
The satellite would escape the planet.

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